大一微积分泰勒公式证明题?
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对∀x∈[0,2],将f(t)在t=x点处泰勒展开
f(t)=f(x)+f'(x)*(t-x)+f''(ξ)/2*(t-x)^2,其中ξ介于x和t之间
将t=0代入,f(0)=f(x)+f'(x)*(-x)+f''(ξ1)/2*x^2
将t=2代入,f(2)=f(x)+f'(x)*(2-x)+f''(ξ2)/2*(2-x)^2
两式相减,f(2)-f(0)=2f'(x)+f''(ξ2)/2*(2-x)^2-f''(ξ1)/2*x^2
f'(x)=f(2)/2-f(0)/2+f''(ξ1)/4*x^2-f''(ξ2)/4*(2-x)^2
|f'(x)|<=|f(2)|/2+|f(0)|/2+|f''(ξ1)|/4*x^2+|f''(ξ2)|/4*(2-x)^2
<=1/2+1/2+(1/4)*x^2+(1/4)*(2-x)^2
=1+(1/4)*(4-4x+2x^2)
=(1/2)*(x^2-2x+4)
=(1/2)*[(x-1)^2+3]
=(1/2)*(x-1)^2+3/2
<=2
f(t)=f(x)+f'(x)*(t-x)+f''(ξ)/2*(t-x)^2,其中ξ介于x和t之间
将t=0代入,f(0)=f(x)+f'(x)*(-x)+f''(ξ1)/2*x^2
将t=2代入,f(2)=f(x)+f'(x)*(2-x)+f''(ξ2)/2*(2-x)^2
两式相减,f(2)-f(0)=2f'(x)+f''(ξ2)/2*(2-x)^2-f''(ξ1)/2*x^2
f'(x)=f(2)/2-f(0)/2+f''(ξ1)/4*x^2-f''(ξ2)/4*(2-x)^2
|f'(x)|<=|f(2)|/2+|f(0)|/2+|f''(ξ1)|/4*x^2+|f''(ξ2)|/4*(2-x)^2
<=1/2+1/2+(1/4)*x^2+(1/4)*(2-x)^2
=1+(1/4)*(4-4x+2x^2)
=(1/2)*(x^2-2x+4)
=(1/2)*[(x-1)^2+3]
=(1/2)*(x-1)^2+3/2
<=2
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